Question

a hot - air balloon plus cargo has a mass of 1890 kg and a volume of 11,490 m³. the balloon is floating at a constant height of 6.25 m above the ground. what is the density of the hot air in the balloon?

a. 7.24 kg/m³

b. 4.83 kg/m³

c. 5.91 kg/m³

d. 1.12 kg/m³

e. 3.22 kg/m³

Explanation:

Step1: Recall the buoyancy principle

The buoyant force $F_b$ equals the weight of the displaced air, and since the balloon is floating at a constant height, $F_b = mg$. The buoyant force formula is $F_b=
ho_{air}Vg$, where $
ho_{air}$ is the density of the surrounding air (assume standard - density of air $
ho_{air} = 1.29\ kg/m^3$), $V$ is the volume of the balloon, and $g$ is the acceleration due to gravity ($g = 9.8\ m/s^2$). Also, $mg=
ho_{hot - air}Vg+ m_{cargo}g$.

Step2: Rearrange the equation for density of hot - air

We can rewrite $mg=
ho_{hot - air}Vg+ m_{cargo}g$ as $
ho_{hot - air}=\frac{m - m_{cargo}}{V}$. Given $m = 1890\ kg$ and $V = 11490\ m^3$.
\[

$$\begin{align*} ho_{hot - air}&=\frac{1890}{11490}\\ &\approx1.12\ kg/m^3 \end{align*}$$

\]

Answer:

d. $1.12\ kg/m^3$